Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{y - 9}{y^2 - 18y + 81} \times \dfrac{3y - 27}{4y + 16} $
Explanation: First factor the quadratic. $n = \dfrac{y - 9}{(y - 9)(y - 9)} \times \dfrac{3y - 27}{4y + 16} $ Then factor out any other terms. $n = \dfrac{y - 9}{(y - 9)(y - 9)} \times \dfrac{3(y - 9)}{4(y + 4)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (y - 9) \times 3(y - 9) } { (y - 9)(y - 9) \times 4(y + 4) } $ $n = \dfrac{ 3(y - 9)(y - 9)}{ 4(y - 9)(y - 9)(y + 4)} $ Notice that $(y - 9)$ appears twice in both the numerator and denominator so we can cancel them. $n = \dfrac{ 3(y - 9)\cancel{(y - 9)}}{ 4\cancel{(y - 9)}(y - 9)(y + 4)} $ We are dividing by $y - 9$ , so $y - 9 \neq 0$ Therefore, $y \neq 9$ $n = \dfrac{ 3\cancel{(y - 9)}\cancel{(y - 9)}}{ 4\cancel{(y - 9)}\cancel{(y - 9)}(y + 4)} $ We are dividing by $y - 9$ , so $y - 9 \neq 0$ Therefore, $y \neq 9$ $n = \dfrac{3}{4(y + 4)} ; \space y \neq 9 $